Zoran Skoda twistsym

Low order Lie algebra type Hopf algebroid of nc phase space of Lie type

$\phi^\nu_\mu = \delta^\nu_\mu + \frac{1}{2}\mathcal{C}^\nu_\mu + \frac{1}{12}(\mathcal{C}^2)^\nu_\mu - \frac{1}{720}(\mathcal{C}^4)^\nu_\mu + O(C^6)$

$L_\mu^\nu = x_\mu\partial^\nu$, $\mathcal{C}^\mu_\nu = \mathcal{C}^\mu_{\nu\lambda}\partial^\lambda$.

Normal ordered form of $\mathcal{F}^{-1}$ up to the third order

$\mathcal{F}^{-1} = 1\otimes 1 + \frac{1}{2} x_\mu\partial^\nu\otimes \mathcal{C}^\mu_\nu + \frac{x_\mu}{12}\left[ \partial^\nu\otimes (\mathcal{C}^2)^\mu_\nu + (\mathcal{C}^2)^\mu_\nu\otimes\partial^\nu \right] + \frac{x_\mu x_{\mu'}}{8}\partial^\nu\partial^{\nu'}\otimes\mathcal{C}^\mu_\nu\mathcal{C}^{\mu'}_{\nu'}$
$-\frac{x_\mu}{24}\partial^\nu\mathcal{C}^\sigma_\rho\otimes\mathcal{C}^\rho_\nu\mathcal{C}^\mu_\sigma +\frac{x_\mu x_{\mu'}}{24}\left[\partial^\nu\partial^{\nu'}\otimes\mathcal{C}^\mu_\nu(\mathcal{C}^2)^{\mu'}_{\nu'} - \mathcal{C}^\mu_\nu(\mathcal{C}^2)^{\mu'}_{\nu'} \otimes\partial^\nu \partial^{\nu'}\right]$

Up to the second order in $C$-s, and with the undeformed exchange rules taken into account, we can write the symmetrized version, $ln(\mathcal{F}^{-1}_S)$, as

$\frac{1}{4}(L_\mu^\nu\otimes\mathcal{C}^\mu_\nu-\mathcal{C}^\mu_\nu\otimes L_\mu^\nu) -\frac{1}{96} (L_\mu^\nu\otimes(\mathcal{C}^2)^\mu_\nu + (\mathcal{C}^2)^\mu_\nu\otimes L_\mu^\nu) + \frac{1}{96} (L_\mu^\nu\mathcal{C}^\mu_\lambda\otimes \mathcal{C}^\lambda_\nu+\mathcal{C}^\lambda_\nu\otimes L_\mu^\nu\mathcal{C}^\mu_\lambda)$

Note that for $N = 0,1,2,...$

$\partial^\nu\partial^{\nu'}\otimes(\mathcal{C}^N)^\mu_\nu\mathcal{C}^{\mu'}_{\nu'} = -\partial^\nu \mathcal{C}^{\mu'}_{\nu'} \otimes \partial^{\nu'}(\mathcal{C}^N)^\mu_\nu$

For the right polarized form $\mathcal{F}^{-1}_R = exp(-\partial^\alpha\otimes x_\alpha) exp(\partial^\lambda\otimes\hat{x}_\lambda)$, using BCH formula up to third order we get

$ln(\mathcal{F}^{-1} = \partial^\alpha\otimes(\hat{x}_\alpha-x_\alpha) - \frac{1}{2} \partial^\alpha\partial^\beta\otimes [x_\alpha,\hat{x}_\beta] +\frac{1}{12}\partial^\alpha\partial^\beta\partial^\gamma\otimes [\hat{x}_\alpha+x_\alpha,[x_\beta,\hat{x}_\gamma]],$

where for the third term we used the fact that $\partial^\beta\partial^\gamma$ is symmetric in $\beta\leftrightarrow\gamma$ hence we renamed the indices.

Now, up to the third order,

$[x_\alpha,\hat{x}_\beta] = [x_\alpha,x_\rho(\delta^\rho_\beta + \frac{1}{2}\mathcal{C}^\rho_\beta + \frac{1}{12}(\mathcal{C}^2)^\rho_\beta)] = -\frac{1}{2} C^\rho_{\alpha\beta} - \frac{1}{12} C_{\beta\alpha}^\sigma \mathcal{C}^\rho_\sigma - \frac{1}{12}\mathcal{C}^\sigma_\beta C^\rho_{\sigma\alpha}$

where the first two terms are antisymmetric and will drop out after contracting with $\partial^\alpha\partial^\beta$. Thus

$-\frac{1}{2}\partial^\alpha\partial^\beta\otimes [x_\alpha,\hat{x}_\beta] = -\frac{1}{24}(\mathcal{C}^2\otimes x_\rho\partial^\lambda)$

Similarly,

$\frac{1}{12}\partial^\alpha\partial^\beta\partial^\gamma\otimes [\hat{x}_\alpha+x_\alpha,[x_\beta,\hat{x}_\gamma]] = \frac{1}{288}\partial^\alpha\partial^\beta\partial^\gamma\otimes [x_\tau\mathcal{C}^\tau_\alpha,x_\rho\mathcal{C}^\sigma_\gamma C^\rho_{\beta\gamma}]$

as other terms drop out by the symmetry. The commutator gives two terms which are equal and of opposite sign, after renaming the indices and contraction with the tensor factor $\partial^\alpha\partial^\beta\partial^\gamma$:

$\frac{1}{288}\partial^\alpha\partial^\beta\partial^\gamma\otimes x_\rho(C^\rho_{\alpha\tau}\mathcal{C}^\sigma_\gamma C^\tau_{\beta\sigma} -\mathcal{C}^\tau_\alpha C^\sigma_{\gamma\tau} C^\rho_{\beta\sigma}) = \frac{1}{288}((\mathcal{C}^2)^\rho_\sigma\partial^\gamma\otimes x_\rho\mathcal{C}^\sigma_\gamma - (\mathcal{C}^2)^\rho_\tau\partial^\gamma \otimes x_\rho\mathcal{C}^\tau_\alpha)=0$

Thus, the contribution in the third order is zero and all up to third order we get that the right polarized twist has exponent

$-\frac{1}{2}\mathcal{C}^\rho_\nu\otimes L^\nu_\rho -\frac{1}{12}\mathcal{C}^\sigma_\lambda\otimes L^\lambda_\rho \mathcal{C}^\rho_\sigma -\frac{1}{24}(\mathcal{C}^2)^\nu_\rho\otimes L_\rho^\nu$

and for the left polarized twist $\mathcal{F}^{-1}_L = exp(-x_\rho\otimes\partial^\rho)exp(\hat{y}_\alpha\otimes\partial^\alpha)$ the exponent is

$+\frac{1}{2}L^\nu_\rho\otimes\mathcal{C}^\rho_\nu -\frac{1}{12}L^\lambda_\rho \mathcal{C}^\rho_\sigma\otimes\mathcal{C}^\sigma_\lambda -\frac{1}{24}L_\rho^\nu\otimes(\mathcal{C}^2)^\nu_\rho$

and if we add the two exponent and divide by 2 we get different coefficients in the 4 second order terms than above.

Last revised on October 24, 2014 at 14:24:28. See the history of this page for a list of all contributions to it.